Question

A box contains 12 light bulbs, of which 5 are defective. All bulbs look alike and have equal

Answer 1

Selecting at least one defective means you could have both defective or one defective with one not defective.  We will find the probability of each and add them together..

Keep in mind we will draw two lights one at a time, replacing the first draw.  Let's calculate the total outcomes possible:  12 possibilities on the first draw and 12 possibilities on the second so 12(12) = 144 total outcomes possible.

P(2 defective) = $\frac{5}{12} ( \frac{5}{12} ) = \frac{25}{144}$
P(1 defective and one non defective) = $\frac{5}{12} ( \frac{7}{12} )= \frac{35}{144}$  In this case first draw not defective is the same as the second not defective because all we are looking for is to have one defective.
P(1 defective w/1 not, or 2 defective) = $\frac{35}{144} + \frac{25}{144} = \frac{60}{144}$ or about 42%

Answer 2

Answer: 95/144


Explanation:

There are 12 light bulbs.
There are 5 defective bulbs.
There are 7 not defective bulbs.

P (at least one defective) = first one defective or second one defective or both defective.

P (at least one defective) = (5/12)(7/12) + (7/12)(5/12) + (5/12)(5/12) = 95/144